Competitors stand in a circle 2.12 meters in diameter and swing the hammer, spinning one to four times before releasing it.

The world record for men is 86.74 meters, set by Yuriy Sedykh in 1986. For women, the record is 79.42 meters, set last year by Betty Heidler.

Simple, right? Not really. The hammer throw combines strength, balance and timing in an event that requires nearly perfect technique. So how does this make the hammer throw like a particle accelerator?

Here is a picture of the synchrotron at Fermi Lab:

Photo: U.S. Dept. of Energy

I would show you something from CERN, but it’s underground and you can’t see anything.

The goal of high-energy particle physics is to get these particles (like a proton) up to really high speeds and then smash them into something. One way of getting these protons moving really fast would be to just put them in a constant electric field. Constant field means constant force and constant acceleration. Simple, right? Well, simple except that you must have something to make this external electric field, something the proton would get out of pretty quick.

Still, they have linear particle accelerators. They are useful for some things, but they can’t get a particle moving as fast as a synchrotron. The two do essentially the same thing, but the big difference is that after the particle leaves the acceleration part of the synchrotron, it enters a magnetic field that curves it around in a circle so the particle can go through the acceleration part again.

Accelerating the particle in a circle increases the effective distance over which the force acts on the particle. Yes, I have simplified this process, but you get the idea. The same thing happens with the hammer throw. If an athlete tried to simply throw the hammer, do you know what would happen? They’d call it the shot put. And so here is a GRE question: The shot put is to a linear accelerator as the hammer throw is to….

The correct answer would be “synchrotron.”

How Fast Are These Hammers?

This is a little tricky. Let me start with the record for the men’s throw at 86.74 meters. I can calculate the initial speed when thrown if I make two assumptions. First, that the hammer was released at certain angle. Let me pick 45° above the horizontal (although I am sure this wasn’t exactly true). Second, that the air drag on the hammer during its motion is small enough to be ignored. I will check this after I have an estimate for the speed.

So, if the ball is thrown with an initial speed v0 at an angle of 45° above the horizontal, this looks like your plain old projectile motion problem. In these types of problems, the only force on the ball is the gravitational force acting downward. This would give a vertical (I will call it the y-direction) of -9.8 m/s2 and a horizontal acceleration of 0 m/s2. Since I know the two accelerations, I can write the following two kinematic equations.

Two important notes here. First, I have made the assumption that the hammer starts at the location x = 0 meters and y = 0 meters. Second, I put the vertical acceleration as -g where g = 9.8 m/s2. If I know the angle θ, how do I find the velocity? If I take the x-equation and solve for the time, I can substitute that expression into the y-equation and then solve for the velocity. If the ball starts and ends at the same height (essentially true) then:

Now, I can put in values for x = 86.7 meters and θ = 45°. This gives an initial hammer speed of about 29 m/s (or about 65 mph).

What About Air Resistance?

Is it ok to assume there is negligible air resistance? One way to answer this question is to calculate the acceleration of the hammer due to just the air resistance force and compare this to the acceleration due to the gravitational force. The typical model for the magnitude of the air resistance force looks like this:

In case you aren’t familiar with this model, here are some details:

v is the magnitude of the hammer’s speed with respect to the air.

ρ is the density of the air (with a value of about 1.2 kg/m3).

A is the cross sectional area of the object. I will assume the hammer looks just like a sphere. This means the cross sectional area will be that of a circle.

C is the drag coefficient. This depends on the shape of the object. A good estimate for a sphere would be about 0.47 (no units).

I can use a velocity of 29 m/s, but what about the radius? The official rules state the length of the hammer, but not the radius of the ball at the end. I wonder if the mass at the end even has to be round? So, let me just say that it is made of steel with a density of about 7800 kg/m3. If the mass of the ball is about 7.2 kg (I took off a small bit for mass of the cable), then this would give a ball volume of 9.2 x 10-4 m3. Assuming it is a sphere, this would give a radius of 6 cm. Now, I can put these values into the air resistance model and get a maximum force of 2.7 Newtons. This would cause an acceleration (due just to the air resistance) of 0.37 m/s2. That’s pretty small compared to the vertical acceleration. I don’t think it is such a terrible idea to ignore the air resistance.

How Does This Even Work?

Now we are getting somewhere. How do you make a ball go faster by swinging it around in a circle? Honestly, I am not quite sure. This means it is experiment time. Step 1: Get daughter to swing ball around outside. Step 2: Record motion by standing on top of swing set (for overhead view). Step 3: video analysis.

Here is the video if you are interested.

Essentially, the string exerts a force on the ball. There are two things this force can do: It can change the speed of the ball or it can change the direction of the motion of the ball. Let me show two shots of this motion. In this first shot, the force is pulling partly in the same direction as the velocity of the ball.

When there is a force in the same direction as the velocity, it will cause the ball to accelerate. When the force is only perpendicular to the direction of the velocity, the force will only cause the ball to change directions. Here is an example of that part of the motion.

You can’t always be pulling the ball in the same direction that it moves or it would “get away from you.” Also, you can’t just pull on the ball perpendicular to the way it is moving or it would never increase in speed.

Moving back to the hammer throw, I suspect essentially the same thing happens. Yes, the person also moves forward during the throw, but I suspect this motion isn’t exactly essential.

Bonus Calculation: Distance Dependence on Launch Angle

Just how critical is this launch angle? If the assumptions of no air resistance and ending at the same height as the start are valid enough, then I can make a plot. This is a plot of the hammer distance as a function of launch angle with an initial speed of 29 m/s.

So you can see that by getting the launch angle off by 5° would shorten your throw. Instead of 86 meters, you would just get around 84 meters. Of course if you launch at 30°, you will take around 10 meters off your range.